[seqfan] Re: (b^k-1)/(b-1) == 1 (mod k)

[Max Alekseyev]

Thomas,

It can be seen that a(n) <= smallest Carmichael number with all prime
factors >= n.
In particular,
a(8..13) <= 29341
a(14..17) <= 162401
a(18..41) <= 252601

I won't be surprised if the equality holds in all cases.

Regards,
Max


On Tue, Apr 24, 2018 at 3:42 AM, Tomasz Ordowski <tomaszordowski at gmail.com>
wrote:

> Dear SeqFan!
>
> Let a(n) be the smallest composite k such that
>
>          b^k == b (mod (b-1)k)
>
> for all integer bases 2 <= b <= n = 2, 3, 4, 5, ...
>
> Conjecture: for n > 2, a(n) is a Carmichael number. Yes?
>
> Can also consider the congruence b^(k-1) == 1 (mod (b-1)k).
>
> Maybe someone will be interested in these sequences.
>
> Please give me some initial terms.
>
> Best regards,
>
> Thomas
>
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